Finding the integration by using residues Hi I am trying to find the following integral
$$\int_{|z|=3}\frac{z^{17}}{(z^2+2)^3(z^3+3)^4}dz$$
There are 5 pole of different order and the calculation is very tedious.
I was wondering if there's any shorter way to find the integral by using residues.
Any help would be highly appreciated. Thanks in advance.
 A: Inasmuch as all of the residues of $f(z)=\frac{z^{17}}{(z^2+2)^3(z^3+3)^4}$ are on the open disk $|z|<3$, we can use the suggestion by @DanielFischer,  Proceesing, we have
$$\begin{align}
\oint_{|z|=3}f(z)\,dz&=-2\pi i \text{Res}(f(z),z=\infty)\\\\
&=2\pi i\text{Res} \left( \frac1{z^2}f\left(\frac{1}{z}\right), z=0\right)\\\\
&=2\pi i\lim_{z\to 0}\frac{1}{(1+2z^2)^3(1+3z^3)^4}\\\\
&=2\pi i
\end{align}$$

Alternatively, we can deform the contour $|z|=3$ to the contour $|z|=R$ where $R$ is chosen large enough to enclose all poles.  Then, we have
$$\begin{align}
\oint_{|z|=3}f(z)\,dz&=\oint_{|z|=R}f(z)\,dz\\\\
&=\int_0^{2\pi} \frac{R^{17}e^{i17\phi}}{R^{18}e^{i18\phi}\left(1+\frac{2}{R^2e^{i2\phi}}\right)^3\left(1+\frac{3}{R^3e^{i3\phi}}\right)^4}\,iRe^{i\phi}\,d\phi\\\\
&=i\int_0^{2\pi}\left(1+O\left(\frac{1}{R^2e^{i2\phi}}\right)\right)\,d\phi\\\\
&=i2\pi
\end{align}$$
since $\int_0^{2\pi}e^{im\phi}\,d\phi=0$ for $m\ne 0$.  
Note, that we could also have simply let $R\to \infty$ and arrived at the same result without exploiting the fact that $\int_0^{2\pi}e^{im\phi}\,d\phi=0$ for $m\ne 0$.
